Bunuel wrote:
A “descending number” is a three-digit number, such that the units digit is less than the tens digits and the tens digit is less than the hundreds digit. What is the probability that a three-digit number chosen at random is a “descending number”?
(A) 3/25
(B) 2/9
(C) 2/15
(D) 1/9
(E) 1/10
Kudos for correct solution.
The total 3 digit numbers can be formed by FCP i.e = 9*10*10 (we have digit from 0 -9 and there are no restriction then the hundred digit can be from 1-9,as it cannot start from 0 , tens digit can be 0-9 and units can also be 0-9).
Now we have to choose the three digit numbers in descending order -
We can leave out 100 - 200 = as they donot form in descending order
Next from 200 to 200 = we have only one number =210
Next from 300 to 400 = we have 3 nos = 310,320,321
Next from 400 to 500 = we have 6 nos = 410,420,421,430,431,432
Now we find that it is following a pattern and i.e
200 = 1
300 = 3 (2+1 from previous possible ways , 2 is the hundred unit )
400 = 6 (3+3)
500 = 10 (4+6,similarly 4 is the hundred digit and 6 in the value from the previous)
600 = 15(5+10)
700 = 21 (6+15)
800 = 28 (7+21)
900 = 36(8+28)
Therefore no. of possible ways = 36 + 28 + 21 + 15 + 10 + 6 + 3 + 1 =120
Therefore \(probability = \frac{120}{900} = \frac{2}{15}\)
_________________
If you found this post useful, please let me know by pressing the Kudos Button
Rules for Posting https://greprepclub.com/forum/rules-for ... -1083.html